CS276 Lecture 13: Pseudorandom Generators

Scribed by Siu-On Chan


Today we complete the proof that it is possible to construct a pseudorandom generator from a one-way permutation.

1. Pseudorandom Generators from One-Way Permutations

Last time we proved the Goldreich-Levin theorem.

Theorem 1 (Goldreich and Levin) Let {f: \{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} be a {(t,\epsilon)}-one way permutation computable in time {r\leq t}. Then the predicate {x,r \mapsto \langle x,r \rangle} is {(\Omega( t \cdot \epsilon^2 \cdot n^{-O(1)} , 3\epsilon)} hard core for the permutation {x,r \mapsto f(x),r}.

A way to look at this result is the following: suppose {f} is {(2^{\Omega(n)},2^{-\Omega(n)})} one way and computable in {n^{O(1)}} time. Then {\langle x,r \rangle} is a {(2^{\Omega(n)},2^{-\Omega(n)})} hard-core predicate for the permutation {x,r \rightarrow f(x),r}.

From now on, we shall assume that we have a one-way permutation {f: \{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} and a predicate {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}} that is {(t,\epsilon)} hard core for {f}.

This already gives us a pseudorandom generator with one-bit expansion.

Theorem 2 (Yao) Let {f:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} be a permutation, and suppose {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}} is {(t,\epsilon)}-hard core for {f}. Then the mapping

\displaystyle  x \mapsto P(x),f(x)

is {(t-O(1), \epsilon)}-pseudorandom generator mapping {n} bits into {n+1} bits.

Note that {f} is required to be a permutation rather than just a function. If {f} is merely a function, it may always begin with 0 and the overall mapping would not be pseudorandom.

For the special case where the predicate {P} is given by Goldreich-Levin, the mapping would be

\displaystyle  x \mapsto \langle x,r \rangle,f(x),r

Proof: Suppose the mapping is not {(t-2,\epsilon)}-pseudorandom. There is an algorithm {D} of complexity {\leq t-2} such that

\displaystyle   \left|\mathop{\mathbb P}_{x\sim\{0,1\}^n} [D(P(x)f(x))=1] - \mathop{\mathop{\mathbb P}_{b\sim\{0,1\}}}_{x\sim\{0,1\}^n} [D(bf(x))=1]\right| > \epsilon \ \ \ \ \ (1)

where we have used the fact that since {f} is permutation, {f(x)} would be a uniformly random element in {\{0,1\}^n} when {x} is such.

We will first remove the absolute sign in (1). The new inequality holds for either {D} or {1-D} (i.e. the complement of {D}), and they both have complexity at most {t-1}.

Now define an algorithm {A} as follows.

On input {y=f(x)}, pick a random bit {r\sim\{0,1\}}. If {D(r,y)=1}, then output {r}, otherwise output {1-r}.

Algorithm {A} has complexity at most {t}. We claim that

\displaystyle  \mathop{\mathbb P}_{x\sim\{0,1\}^n} [A(f(x))=P(x)]> \frac12+\epsilon

so {P(\cdot)} is not {(t,\epsilon)}-hard core.

To make explicit the dependence of {A} on {r}, we will denote by {A_r(f(x))} the fact that {A} picks {r} as its random bit.

To prove the claim, we expand

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb P}_{x,r} [A_r(f(x))=P(x)] &= & \mathop{\mathbb P}_{x,r} [A_r(f(x))=P(x)\mid r=P(x)]\mathop{\mathbb P}[r=P(x)] + \\ && \mathop{\mathbb P}_{x,r} [A_r(f(x))=P(x)\mid r\neq P(x)]\mathop{\mathbb P}[r\neq P(x)] \\ \end{array}

Note that {\mathop{\mathbb P}[r=P(x)]=\mathop{\mathbb P}[r\neq P(x)]=1/2} no matter what {P(x)} is. The above probability thus becomes

\displaystyle   \frac12 \mathop{\mathbb P}_{x,r} [D(rf(x))=1\mid r=P(x)] + \frac12 \mathop{\mathbb P}_{x,r} [D(rf(x))=0\mid r\neq P(x) ] \ \ \ \ \ (2)

The second term is just {\frac12 - \frac12 \mathop{\mathbb P}_{x,r} [D(rf(x))=1\mid r\neq P(x) ]}. Now we add to and subtract from (2) the quantity {\frac12 \mathop{\mathbb P}_{x,r} [D(rf(x))=1\mid r=P(x) ]}, getting

\displaystyle  \frac12 + \mathop{\mathbb P}_{x,r} [D(rf(x))=1\mid r=P(x)] -

\displaystyle  \left(\frac12 \mathop{\mathbb P}[D(rf(x))=1\mid r=P(x)] + \right.

\displaystyle  \left.\frac12 \mathop{\mathbb P}[D(rf(x))=1\mid r\neq P(x)]\right)

The expression in the bracket is {\mathop{\mathbb P}[D(rf(x))=1]}, and by our assumption on {D}, the whole expression is more than {\frac12 + \epsilon}, as claimed. \Box

The main idea of the proof is to convert something that distinguishes (i.e. {D}) to something that outputs (i.e. {A}). {D} helps us distinguish good answers and bad answers.

We will amplify the expansion of the generator by the following idea: from an {n}-bit input, we run the generator to obtain {n+1} pseudorandom bits. We output one of those {n+1} bits and feed the other {n} back into the generator, and so on. Specialized to the above construction, and repeated {k} times the mapping becomes

\displaystyle   G_k (x) := P(x), P(f(x)), P(f(f(x)), \ldots, P(f^{(k-1)} (x), f^{(k)} (x) \ \ \ \ \ (3)

This corresponds to the following diagram where all output bits lie at the bottom.

Theorem 3 (Blum-Micali) Let {f:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} be a permutation, and suppose {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}} is {(t,\epsilon)}-hard core for {f} and that {f,P} are computable with complexity {r}.

Then {G_k : \{ 0,1 \}^n \rightarrow \{ 0,1 \}^{n+k}} as defined in (3) is {(t-O(rk), \epsilon k)}-pseudorandom.

Proof: Suppose {G_k} is not {(t-O(rk),\epsilon k)}-pseudorandom. Then there is an algorithm {D} of complexity at most {t-O(rk)} such that

\displaystyle  \left| \mathop{\mathbb P}_{x\sim\{0,1\}^n} [D(G_k(x))=1]-\mathop{\mathbb P}_{z\sim\{0,1\}^{n+k}} [D(z)=1] \right| > \epsilon k

We will then use the hybrid argument. We will define a sequence of distributions {H_0,\dots,H_k}, the first is {G_k}‘s output, the last is uniformly random bits, and every two adjacent ones differ only in one invocation of {G}.

More specifically, define {H_i} to be the distribution where we intercept the output of the first {i} copies of {G}‘s, replace them with random bits, and run the rest of {G_k} as usual (see the above figure in which blue lines represent intercepted outputs). Then {H_0} is just the distribution of the output of {G_k}, and {H_k} is the uniform distribution, as desired. Now

\displaystyle  \begin{array}{rcl}  \epsilon k &< & \left|\mathop{\mathbb P}_{z\sim H_0} [D(z)=1]-\mathop{\mathbb P}_{z\sim H_k} [D(z)=1]\right| \\ &= & \left|\sum_{i=0}^{k-1}\left(\mathop{\mathbb P}_{z\sim H_i} [D(z)=1]-\mathop{\mathbb P}_{z\sim H_{i+1}} [D(z)=1] \right) \right| \end{array}

So there is an {i} such that

\displaystyle  \left|\mathop{\mathbb P}_{z\sim H_i} [D(z)=1]-\mathop{\mathbb P}_{z\sim H_{i+1}}[D(z)=1]\right| > \epsilon

In both {H_i} and {H_{i+1}}, the first {i} bits {r_1,\dots,r_i} are random. We now define a new algorithm {D'} that takes as input {b,y} and has output distribution {H_i} or {H_{i+1}} in two special cases: if {b,y} are drawn from {P(x),f(x)}, then {D'} has output distribution {H_i}; if {b,y} are drawn from (random bit),{f(x)}, then {D'} has output distribution {H_{i+1}}. In other words, if {b,y} are {P(x),f(x)}, {D'} should output

\displaystyle  r_1,\dots,r_i,P(x),P(f(x)),\dots,P(f^{(k-i-1)}(x)),f^{(k-i)}(x)

If {b,y} are (random bit),{f(x)}, {D'} should output

\displaystyle  r_1,\dots,r_i,r_{i+1},P(f(x)),\dots,P(f^{(k-i-1)}(x)),f^{(k-i)}(x)

This suggests that {D'} on input {b,y} should pick random bits {r_1,\dots,r_i} and output {r_1,\dots,r_i,b,P(y),\dots,P(f^{(k-i-2)}(y)),f^{(k-i-1)}(y)}.

We have

\displaystyle  \begin{array}{rcl}  & & \left| \mathop{\mathbb P}_{x\sim\{0,1\}^n}[D'(P(x)f(x))=1]-\mathop{\mathbb P}_{z\sim\{0,1\}^{n+1}}[D'(z)=1] \right| \\ &= & \left| \mathop{\mathbb P}_{x\sim H_i} [D'(x)=1] - \mathop{\mathbb P}_{x\sim H_{i+1}} [D'(x)=1] \right| \\ &> & \epsilon \end{array}

and {P(\cdot)} is not {(t,\epsilon)}-hard core. \Box

Thinking about the following problem is a good preparation for the proof the main result of the next lecture.

Exercise 1 (Tree Composition of Generators) Let {G:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^{2n}} be a {(t,\epsilon)} pseudorandom generator computable in time {r}, let {G_0(x)} be the first {n} bits of the output of {G(x)}, and let {G_1(x)} be the last {n} bits of the output of {G(x)}.

Define {G' : \{ 0,1 \}^n \rightarrow \{ 0,1 \}^{4n}} as

\displaystyle  G' (x) = G(G_0(x)), G( G_1 (x))

Prove that {G'} is a {(t-O(r),3\epsilon)} pseudorandom generator.

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